# Find (x/y)^3?

If 1/x + 1/y = 1/(x+y) find (x/y)^3

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• 8 months ago

(1/x) + (1/y) = 1/(x + y)

[(1/x) + (1/y)] * x = [1/(x + y)] * x

1 + (x/y) = x/(x + y)

(x/y) + 1 = (x/y)/[(x + y)/y]

(x/y) + 1 = (x/y)/[(x/y) + 1]

Put a = x/y:

a + 1 = a/(a + 1)

(a + 1)² = a

a² + 2a +1 = a

a² + a + 1 = 0

(a - 1)(a² + a + 1) = 0

a³ - 1 = 0

a³ = 1

Hence, (x/y)³ = 1

• JOHN
Lv 7
8 months ago

Set u = x/y

x(1/x + 1/y) = x/(x + y)

1 + u = 1/(1 + 1/u) = u/(1 + u)

(u + 1)² = u

u² + 2u + 1 = u

u² + u + 1 = 0

u³ - 1 = (u - 1)(u² + u + 1) = 0

u³ = (x/y)³ = 1.

• 8 months ago

xy + y^2 + x^2 + xy = xy =>

x^2 + xy + y^2 = 0 =>

x = -y/2 +/- (yi/2)*sqrt(3) =>

x/y = -1/2 +/- i*sqrt(3)/2 =>

(x/y)^3 = -1.

• BigBird8 months agoReport

{1e^[(+/-)(2/3)(pi)] }^3 = e^2(pi) = +1

• 8 months ago

1/x + 1/y = 1/(x+y)

(x+y)/(xy) = 1/(x+y)

(x+y)² = xy

x² + xy + y² = 0

x = -y/2 ± √(y²-4y²)/2 = -y/2 ± i y√3/2

x/y = -1/2 ± i √3/2

(x/y)³ = (-½ ± i ½√3)³ = 1

• BigBird8 months agoReport

{1e^[(+/-)(2/3)(pi)] }^3 =e^[2(pi)] = +1